Matrices

2x-y+2z=2
x-2y+λz=-4
x+λy+z=4
For no solution:
D=|2, -1,2; 1, -2, λ 1,1,1|=0
⇒ 2 (-2-λ²)+1 (1-λ)+2 (λ+2)=0
⇒ -2λ²+λ+1=0
⇒ λ=1, -1/2
D? =|2, -1,2; -4, -2, λ 4,1,1|
=2 (-2-λ)+1 (-4-4λ)+2 (-4+8)
=2 (1+λ) which is not equal to zero for λ=1, -1/2

|A|≠0
For (P): A≠I?
So, A = [1 0; 0 1] or [1; 0 1] or [1 0; 1]
or [1; 1 0]
So (P) is false.
A = [1 0; 1 0] or [1; 0 1] or [1 0; 1]
⇒ tr (A)=2
⇒ Q is true

Δ = |1,1, -1; 1,2, α 2, -1,1| = 1 (2+α)-1 (1-2α)-1 (-1-4) = 2+α+2α-1+5 = 3α+6=0 ⇒ α=-2.
Δ? = |2,1, -1; 1,2, α β, -1,1| = 2 (2+α)-1 (1-αβ)-1 (-1-2β) = 4+2α-1+αβ+1+2β = 4+2α+αβ+2β=0.
4-4-2β+2β=0. This holds.
Δ? = |1,2, -1; 1,1, α 2, β,1| = 1 (1-αβ)-2 (1-2α)-1 (β-2) = 1-αβ-2+4α-β+2 = 1+4α-αβ-β=0.
1-8+2β-β=0 ⇒ -7+β=0 ⇒ β=7.
α+β = -2+7 = 5.

Given x + 2y – 3z = a

2x + 6y – 11z = b

x – 2y + 7z = c

Here $\begin{array}{cc}\hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill \end{array}\begin{array}{cc}\hfill \begin{array}{l}\\ \end{array}\hfill & \hfill \hfill \end{array}$

$\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 2\hfill & \hfill 6\hfill & \hfill -11\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 7\hfill \end{array}\right|\begin{array}{cc}\hfill \begin{array}{l}=\left(42-22\right)-2\left(14+11\right)-3\left(-4-6\right)\\ \end{array}\hfill & \hfill =20-50+30=0\hfill \end{array}$

For infinite solution ${}_{}{}_{}{}_{}$

20a – 8b – 4c = 0 5a = 2b + c

  $A^2=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$A^3=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^3\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$
$A^4\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^3\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

Similarly we get A¹⁹ =   $=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^{19}\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\&A^{20}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^{20}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

=  $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\Rightarrow 1+\alpha +\beta =1gives\alpha +\beta =0........\left(i\right)$

$2^{20}+\left(2^{19}-2\right)\alpha =4from\left(i\right)$

$\Rightarrow \alpha =\frac{4-2^{20}}{2^{19}-2}=\frac{4\left(1-2^{18}\right)}{-2\left(1-2^{18}\right)}=-2$

So, β = 2

Hence β - α = 4

$A={\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)}_{3*3}$

 B is a matrix of same order with entries from {1,2,3,4,5}. and satisfying AB = BA.

$LetB=\left(\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right)$           

$\therefore \left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$

$a_{21}=a_{12},a_{22}=a_{11},a_{23}=a_{13},a_{31}=a_{32}$

$\therefore$ there exist only 5 distinct entries in the matrix B so that possible case = 5⁵ = 3125

  $lo{g}_{9^{\frac{1}{2}}}x+lo{g}_{9^{\frac{1}{3}}}x+lo{g}_{9^{\frac{1}{4}}}x+.......+lo{g}_{9^{\frac{1}{22}}}x=504$

=> $2lo{g}_{9}x+3lo{g}_{9}x+4lo{g}_{9}x+........+22lo{g}_{9}x=504$

=> (1 + 2 + 3 + .+ 22) log₉ x – log₉ x = 504 Þ x = 81

$I-A^3-3A+3A^2=I-A^3$

=> 3A² – 3A = 0

=> 3A (A – I) = 0

=>A² = A

$\left[\begin{array}{cc}\hfill a^2\hfill & \hfill ab+bd\hfill \\ \hfill 0\hfill & \hfill d^2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill d\hfill \end{array}\right]$    

Total number of ways = 8

$A=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]then{A}^2=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2k\hfill & \hfill -2\hfill \\ \hfill -k\hfill & \hfill 2k+1\hfill \end{array}\right]$

$A^3=\left[\begin{array}{cc}\hfill 2k\hfill & \hfill -2\hfill \\ \hfill -k\hfill & \hfill 2k+1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill -2k\hfill & \hfill 4k+2\hfill \\ \hfill 2k^2+k\hfill & \hfill -4k-1\hfill \end{array}\right]$              

Now, A(A³ + 3l) = 2l gives A³ + 3l = 2A^-1

$\Rightarrow -2k+3=\frac{1}{k}\Rightarrow 2k^2-3k+1=0$    

  $k=\frac{1}{2},1$            

& $4k+2=\frac{2}{k}or2k+1-\frac{1}{k}soor2k^2+k-1=0$

=> $k=\frac{1}{2}$

AA^T = A^TA = l

$A{\left(A^{T}BA\right)}^{2021}{A}^T$              

=  $B^{2021}={\left(I+P\right)}^{2021}=l+2021P=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 2021i\hfill & \hfill 1\hfill \end{array}\right]$

$\Rightarrow {\left(B^{2021}\right)}^{-1}=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -2021i\hfill & \hfill 1\hfill \end{array}\right]$

where P = $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill i\hfill & \hfill 0\hfill \end{array}\right]$